Calculate the concentration of sulfate in a solution made from 65.0 mL of 0.55 M sodium sulfate and 85.0 mL of 1.25 M iron(III) sulfate
Concentration of mixture = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
where C1 --> Concentration of 1 component
V1-->volume of 1 component
C2 --> Concentration of other component
V2-->volume of other component
n1 --> number of particle from 1 molecule of 1st component
= 1 because 1 mol of Na2SO4 gives 1 mol of SO42-
n2 --> number of particle from 1 molecule of 2nd component
= 3 because 1 mol of Fe2(SO4)3 gives 3 mol of SO42-
use:
C = (n1*C1*V1+ n2*C2*V2) / (V1+V2)
C = (1*0.55*65+3*1.25*85)/(65+85)
C = 2.36 M
Answer: 2.36 M
Calculate the concentration of sulfate in a solution made from 65.0 mL of 0.55 M sodium...
Calculate the concentration (M) of sodium ions in a solution made by diluting 30.0 mL of a 0.574 M solution of sodium sulfide to a total volume of 150 mL.
A mixture was made from solutions of sodium sulfate and aluminium sulfate. 313 mL of 0.500 M Na2SO4 solution and 187 mL of 0.300 M Al2(SO )3 solution were mixed together. Calculate the concentration of SO 2-ions in the final solution? MgSi(s) + 4H2O(0) - 2Mg(OH)2(aq) + SiH.(g) How much magnesium hydroxide will be obtained if 50,0 g of Mg Si and 25.0 g of water are reacted?
1. Calculate the concentration (M) of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of sodium sulfide to a total volume of 250. mL 2. Which element is oxidized in the reaction below? Fe+2 + H+ + Cr2O7-2 --> Fe+3 + Cr+3 + H2O 3. How many moles of BaCl2 are formed in the neutralization of 393ml of 0.171 M Ba(OH)2 with aqueous HCL?
Calculate the molarity of sodium ion in a solution made by mixing 2.81 mL of 0.169 M sodium chloride with 475.00 mL of 9.36 X 10M sodium sulfate. Assume volumes are additive M solution
Calculate the molarity of sodium ion in a solution made by mixing 5.16 mL of 0.799 M sodium chloride with 475.00 mL of 5.57 × 10−2 M sodium sulfate. Assume volumes are additive.
Calculate the molarity of sodium ion in a solution made by mixing 5.89 mL of 0.130 M sodium chloride with 475.00 mL of 7.27 × 10−2 M sodium sulfate. Assume volumes are additive.
[18] Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (NagSO4) in enough water to form 125 mL of solution. A) .125 M B) 1.32 M C) 132 M D) 125M
Solid barium acetate is slowly added to 150 ml of a 0.395 M sodium sulfate solution until the concentration of barium ion is 0.0319 M. The percent of sulfate ion remaining in solution is %
65.0 mL sample of a 0.102 M potassium sulfate solution is mixed with 38.5 mL of a 0.122 M lead(II) acetate solution and the following precipitation reaction occurs: K2SO4(aq)+Pb(C2H3O2)2(aq)→2KC2H3O2(aq)+PbSO4(s) The solid PbSO4 is collected, dried, and found to have a mass of 0.993 g . a. Define the theoretical yield b. Define the percent yield.
Solid sodium sulfate is slowly added to 150 mL of a 0.0595 M barium acetate solution. The concentration of sulfate ion required to just initiate precipitation is M